Maxima and Minima Trig Problem

**astartleddeer** · July 20th 2012, 07:24 AM

Hello,

I'm having difficulty in understanding where the book has got these angles from:

Question: Find the maximum and minimum values for $y = 3\sin\theta - 4\cos\theta\ in\ the\ range\ \theta\ to\ 2\pi$

$\frac{d}{d\theta} (3\sin\theta - 4\cos\theta) = 4\sin\theta + 3\cos\theta$

$4\sin\theta + 3\cos\theta = R\sin(\theta + \alpha)$

$4\sin\theta + 3\cos\theta = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha$

$R = \sqrt {4^2 + 3^2} = \pm5$

$\alpha = \arctan \frac{3}{4} \approx 36.869^{\circ}$

$\frac {dy}{d\theta} = 5\sin(\theta + 36.869^{\circ}) = \pm 5\ for\ a\ maximum\ or\ minimum\ value$

$\theta \approx 90 - 36.869^{\circ} \approx 53.131^{\circ}$

Now, in my book the answer is 5 at $143^{\circ}8'$ for a maximum value, and -5 at $323^{\circ}8'$ for a minimum value. However, they have subtracted $\alpha$ from $180^{\circ}$ in order to obtain these pair of angles? I've drawn the function, and these values is where $5\sin(\theta + 36.869^{\circ}) = 0$ . As a matter of fact, the maximum value first occurs at $53.131^{\circ}$ and the minimum value at $233.131^{\circ}$ .

I've looked into this a bit deeper, and the only way a maximum or minimum value can occur at these computed angles - is if the first derivative is a negative cosine function.

What have I done wrong?

Thank you for your attention.

**skeeter** · July 20th 2012, 08:10 AM

first off, the problem states the interval of interest is $0 \le \theta < 2\pi$ ... you need to start thinking radians.

$y = 3\sin{\theta} - 4\cos{\theta}$

$\frac{dy}{d\theta} = 3\cos{\theta} + 4\sin{\theta} = 0$

for $\frac{dy}{d\theta} = 0$ , $\cos{\theta}$ and $\sin{\theta}$ will have to have opposite signs ... which tells you $\theta$ will have to be in quadrants II or IV.

$\tan{\theta} = -\frac{3}{4}$

$\theta = \arctan\left(-\frac{3}{4}\right) + \pi$ , a quad II angle

$\theta = \arctan\left(-\frac{3}{4}\right) + 2\pi$ , a quad IV angle

to finish, use the second derivative test for the two angles to determine max/min.

**astartleddeer** · July 20th 2012, 08:29 AM

for $\frac{dy}{d\theta} = 0$ , $\cos{\theta}$ and $\sin{\theta}$ will have to have opposite signs ... which tells you $\theta$ will have to be in quadrants II or IV.

$\tan{\theta} = -\frac{3}{4}$

$\theta = \arctan\left(-\frac{3}{4}\right) + \pi$ , a quad II angle

$\theta = \arctan\left(-\frac{3}{4}\right) + 2\pi$ , a quad IV angle

to finish, use the second derivative test for the two angles to determine max/min.

I don't understand. For $\frac {dy}{d\theta}$ why do we have to give sin and cos opposite signs to equal zero?

If sin and cos are both made positive does this not make $\tan \theta = \frac{-3}{-4} = \frac {3}{4}$ ?

Further, why have you added $\pi$ and $2\pi$ to $\tan = -\frac{3}{4}$ ?

**skeeter** · July 20th 2012, 08:42 AM

Originally Posted by astartleddeer

I don't understand. For $\frac {dy}{d\theta}$ why do we have to give sin and cos opposite signs to equal zero?

3(something) + 4(something else) = 0

what does that tell you about the signs of (something) and (something else) ?

If sin and cos are both made positive does this not make $\tan \theta = \frac{-3}{-4} = \frac {3}{4}$ ?

if sine and cosine have the same sign, then tangent is positive ... the solutions for y' = 0 is where tangent is negative.

Further, why have you added $\pi$ and $2\pi$ to $\tan = -\frac{3}{4}$ ?

you need to research the range of the arctangent function.

**astartleddeer** · July 20th 2012, 09:41 AM

Originally Posted by skeeter

3(something) + 4(something else) = 0

what does that tell you about the signs of (something) and (something else) ?

Then only one of them can be negative?

If sine and cosine have the same sign, then tangent is positive ... the solutions for y' = 0 is where tangent is negative.

Oops. I was meant to say, if sin and cos are both negative then the tangent is positive in the third quadrant?

you need to research the range of the arctangent function.

Oh my goodness! I had mistaken $\pi$ for $90^{\circ}$ ! Yes, I'm with you now. If the tangent is negative, it is $\pi + (-0.64^{c})$ or $2\pi + (-0.64^{c})$ . Second and fourth quadrant respectively.

**astartleddeer** · July 20th 2012, 11:07 AM

Ok let's do this.

$\frac{dy}{d\theta} = 4\sin\theta - 3\cos\theta = 0\ for\ a\ maximum\ or\ minimum\ value$

$R = \sqrt{4^2 + (-3)^2} = \pm 5$

$\theta = \arctan -\frac{3}{4} \approx -0.6435^{c}$

$In\ the\ range\ 0 \leq \theta \leq 2\pi...\theta \approx \pi + (-0.6435^{c}) \approx 2.496^{c}$ and $2\pi + (-0.6435^{c}) \approx 5.639^{c}$

$\frac{d^{2}y}{d\theta^2} = 3\sin\theta + 4\cos\theta$

$\frac{d^{2}y}{d\theta^2}$ is a maximum for $2.496^{c}$

$\frac{d^{2}y}{d\theta^2}$ is a minimum for $5.639^{c}$

Therefore, a minimum value occurs at $-5, 5.639^{c}$ and a maximum value occurs at $5, 2.496^{c}$

**skeeter** · July 20th 2012, 11:15 AM

Originally Posted by astartleddeer

Ok let's do this.

[tex] \frac{dy}{d\theta} = 4\sin\theta - 3\cos\theta = 0\ for\ a\ maximum\ or\ minimum\ value

$R = \sqrt{4^2 + (-3)^2} = \pm 5$

$\theta = \arctan = -\frac{3}{4} \approx -0.6435^{\circ}$

$In\ the\ range\ 0 \leq \theta \leq 2\pi...\theta \approx \pi + (-0.6435^{c}) \approx 2.496^{\circ}$ and $2\pi + (-0.6435^{c}) \approx 5.639^{c}$

$\frac{d^{2}y}{d\theta^2} = 3\sin\theta + 4\cos\theta$

$\frac{d^{2}y}{d\theta^2}$ is a maximum for $2.496^{c}$

$\frac{d^{2}y}{d\theta^2}$ is a minimum for $5.639^{c}$

Therefore, a minimum value occurs at $-5, 5.639^{c}$ and a maximum value occurs at $5, 2.496^{c}$

Please confirm this.

maximum function value of $y = 5$ at $\theta = \arctan\left(-\frac{3}{4}\right) + \pi \approx 2.498$ RADIANS

minimum function value of $y = -5$ at $\theta = \arctan\left(-\frac{3}{4}\right) + 2\pi \approx 5.640$ RADIANS

**astartleddeer** · July 20th 2012, 11:24 AM

I indicated radians with a small c for circular measure

Ok, thank you for your time Skeeter. Done and dusted

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